Bacterial Colony Doubling Problem
Let the initial number of bacteria be \( P_0 \).
Given that the colony doubles every 3 hours, the number of bacteria after 3 hours (\( P_3 \)) is \( 2P_0 \).
The number of bacteria after 6 hours (\( P_6 \)) is \( 2P_3 = 2 \times 2P_0 = 2^2P_0 \).
The number of bacteria after 9 hours (\( P_9 \)) is \( 2P_6 = 2 \times 2^2P_0 = 2^3P_0 \).
The number of bacteria after 12 hours (\( P_{12} \)) is \( 2P_9 = 2 \times 2^3P_0 = 2^4P_0 \).
Thus, after 12 hours, the number of bacteria is \( P_{12} = 2^4P_0 \).
Since \( P_0 = 10 \):
\( P_{12} = 2^4 \times 10 \)
\( P_{12} = 16 \times 10 \)
\( P_{12} = 160 \)
Therefore, the number of bacteria in the colony after 12 hours is 160. The answer is option C.
